Convert month int to month name in Pandas
Question:
I want to transform an integer between 1 and 12 into an abbrieviated month name.
I have a df which looks like:
client Month
1 sss 02
2 yyy 12
3 www 06
I want the df to look like this:
client Month
1 sss Feb
2 yyy Dec
3 www Jun
Most of the info I found was not in python>pandas>dataframe hence the question.
Answers:
You can do this easily with a column apply.
import pandas as pd
df = pd.DataFrame({'client':['sss', 'yyy', 'www'], 'Month': ['02', '12', '06']})
look_up = {'01': 'Jan', '02': 'Feb', '03': 'Mar', '04': 'Apr', '05': 'May',
'06': 'Jun', '07': 'Jul', '08': 'Aug', '09': 'Sep', '10': 'Oct', '11': 'Nov', '12': 'Dec'}
df['Month'] = df['Month'].apply(lambda x: look_up[x])
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
One way of doing that is with the apply method in the dataframe but, to do that, you need a map to convert the months. You could either do that with a function / dictionary or with Python’s own datetime.
With the datetime it would be something like:
def mapper(month):
date = datetime.datetime(2000, month, 1) # You need a dateobject with the proper month
return date.strftime('%b') # %b returns the months abbreviation, other options [here][1]
df['Month'].apply(mapper)
In a simillar way, you could build your own map for custom names. It would look like this:
months_map = {01: 'Jan', 02: 'Feb'}
def mapper(month):
return months_map[month]
Obviously, you don’t need to define this functions explicitly and could use a lambda directly in the apply method.
You can do this efficiently with combining calendar.month_abbr and df[col].apply()
import calendar
df['Month'] = df['Month'].apply(lambda x: calendar.month_abbr[x])
Use strptime and lambda function for this:
from time import strptime
df['Month'] = df['Month'].apply(lambda x: strptime(x,'%b').tm_mon)
Having tested all of these on a large dataset, I have found the following to be fastest:
import calendar
def month_mapping():
# I'm lazy so I have a stash of functions already written so
# I don't have to write them out every time. This returns the
# {1:'Jan'....12:'Dec'} dict in the laziest way...
abbrevs = {}
for month in range (1, 13):
abbrevs[month] = calendar.month_abbr[month]
return abbrevs
abbrevs = month_mapping()
df['Month Abbrev'} = df['Date Col'].dt.month.map(mapping)
Since the abbreviated month names is the first three letters of their full names, we could first convert the Month column to datetime and then use dt.month_name() to get the full month name and finally use str.slice() method to get the first three letters, all using pandas and only in one line of code:
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.month_name().str.slice(stop=3)
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
The calendar module is useful, but calendar.month_abbr is array-like: it cannot be used directly in a vectorised fashion. For an efficient mapping, you can construct a dictionary and then use pd.Series.map:
import calendar
d = dict(enumerate(calendar.month_abbr))
df['Month'] = df['Month'].map(d)
Performance benchmarking shows a ~130x performance differential:
import calendar
d = dict(enumerate(calendar.month_abbr))
mapper = calendar.month_abbr.__getitem__
np.random.seed(0)
n = 10**5
df = pd.DataFrame({'A': np.random.randint(1, 13, n)})
%timeit df['A'].map(d) # 7.29 ms per loop
%timeit df['A'].map(mapper) # 946 ms per loop
Solution 1: One liner
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.strftime('%b')
Solution 2: Using apply()
def mapper(month):
return month.strftime('%b')
df['Month'] = df['Month'].apply(mapper)
Reference:
You can use Pandas month_name() function. Example:
>>> idx = pd.date_range(start='2018-01', freq='M', periods=3)
>>> idx
DatetimeIndex(['2018-01-31', '2018-02-28', '2018-03-31'],
dtype='datetime64[ns]', freq='M')
>>> idx.month_name()
Index(['January', 'February', 'March'], dtype='object')
For more detail visit this link.
Suppose we have a DF like this, and Date is already in DateTime Format:
df.head(3)
value
date
2016-05-19 19736
2016-05-26 18060
2016-05-27 19997
Then we can extract month number and month name easily like this :
df['month_num'] = df.index.month
df['month'] = df.index.month_name()
value year month_num month
date
2017-01-06 37353 2017 1 January
2019-01-06 94108 2019 1 January
2019-01-05 77897 2019 1 January
2019-01-04 94514 2019 1 January
using datetime object methods
I’m surpised this answer doesn’t have a solution using strftime
note, you’ll need to have a valid datetime object before using the strftime method, use pd.to_datetime(df['date_column']) to cast your target column to a datetime object.
import pandas as pd
dates = pd.date_range('01-Jan 2020','01-Jan 2021',freq='M')
df = pd.DataFrame({'dates' : dates})
df['month_name'] = df['dates'].dt.strftime('%b')
dates month_name
0 2020-01-31 Jan
1 2020-02-29 Feb
2 2020-03-31 Mar
3 2020-04-30 Apr
4 2020-05-31 May
5 2020-06-30 Jun
6 2020-07-31 Jul
7 2020-08-31 Aug
8 2020-09-30 Sep
9 2020-10-31 Oct
10 2020-11-30 Nov
11 2020-12-31 Dec
another method would be to slice the name using dt.month_name()
df['month_name_str_slice'] = df['dates'].dt.month_name().str[:3]
dates month_name month_name_str_slice
0 2020-01-31 Jan Jan
1 2020-02-29 Feb Feb
2 2020-03-31 Mar Mar
3 2020-04-30 Apr Apr
4 2020-05-31 May May
5 2020-06-30 Jun Jun
6 2020-07-31 Jul Jul
7 2020-08-31 Aug Aug
8 2020-09-30 Sep Sep
9 2020-10-31 Oct Oct
10 2020-11-30 Nov Nov
11 2020-12-31 Dec Dec
the best way would be to do with month_name() as commented by
Nurul Akter Towhid.
df[‘Month’] = df.Month.dt.month_name()
First you need to strip "0 " in the beginning (as u might get the exception leading zeros in decimal integer literals are not permitted; use an 0o prefix for octal integers)
step1)
def func(i):
if i[0] == '0':
i = i[1]
return(i)
df["Month"] = df["Month"].apply(lambda x: func(x))
Step2:
df["Month"] = df["Month"].apply(lambda x: calendar.month_name(x))
I want to transform an integer between 1 and 12 into an abbrieviated month name.
I have a df which looks like:
client Month
1 sss 02
2 yyy 12
3 www 06
I want the df to look like this:
client Month
1 sss Feb
2 yyy Dec
3 www Jun
Most of the info I found was not in python>pandas>dataframe hence the question.
You can do this easily with a column apply.
import pandas as pd
df = pd.DataFrame({'client':['sss', 'yyy', 'www'], 'Month': ['02', '12', '06']})
look_up = {'01': 'Jan', '02': 'Feb', '03': 'Mar', '04': 'Apr', '05': 'May',
'06': 'Jun', '07': 'Jul', '08': 'Aug', '09': 'Sep', '10': 'Oct', '11': 'Nov', '12': 'Dec'}
df['Month'] = df['Month'].apply(lambda x: look_up[x])
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
One way of doing that is with the apply method in the dataframe but, to do that, you need a map to convert the months. You could either do that with a function / dictionary or with Python’s own datetime.
With the datetime it would be something like:
def mapper(month):
date = datetime.datetime(2000, month, 1) # You need a dateobject with the proper month
return date.strftime('%b') # %b returns the months abbreviation, other options [here][1]
df['Month'].apply(mapper)
In a simillar way, you could build your own map for custom names. It would look like this:
months_map = {01: 'Jan', 02: 'Feb'}
def mapper(month):
return months_map[month]
Obviously, you don’t need to define this functions explicitly and could use a lambda directly in the apply method.
You can do this efficiently with combining calendar.month_abbr and df[col].apply()
import calendar
df['Month'] = df['Month'].apply(lambda x: calendar.month_abbr[x])
Use strptime and lambda function for this:
from time import strptime
df['Month'] = df['Month'].apply(lambda x: strptime(x,'%b').tm_mon)
Having tested all of these on a large dataset, I have found the following to be fastest:
import calendar
def month_mapping():
# I'm lazy so I have a stash of functions already written so
# I don't have to write them out every time. This returns the
# {1:'Jan'....12:'Dec'} dict in the laziest way...
abbrevs = {}
for month in range (1, 13):
abbrevs[month] = calendar.month_abbr[month]
return abbrevs
abbrevs = month_mapping()
df['Month Abbrev'} = df['Date Col'].dt.month.map(mapping)
Since the abbreviated month names is the first three letters of their full names, we could first convert the Month column to datetime and then use dt.month_name() to get the full month name and finally use str.slice() method to get the first three letters, all using pandas and only in one line of code:
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.month_name().str.slice(stop=3)
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
The calendar module is useful, but calendar.month_abbr is array-like: it cannot be used directly in a vectorised fashion. For an efficient mapping, you can construct a dictionary and then use pd.Series.map:
import calendar
d = dict(enumerate(calendar.month_abbr))
df['Month'] = df['Month'].map(d)
Performance benchmarking shows a ~130x performance differential:
import calendar
d = dict(enumerate(calendar.month_abbr))
mapper = calendar.month_abbr.__getitem__
np.random.seed(0)
n = 10**5
df = pd.DataFrame({'A': np.random.randint(1, 13, n)})
%timeit df['A'].map(d) # 7.29 ms per loop
%timeit df['A'].map(mapper) # 946 ms per loop
Solution 1: One liner
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.strftime('%b')
Solution 2: Using apply()
def mapper(month):
return month.strftime('%b')
df['Month'] = df['Month'].apply(mapper)
Reference:
You can use Pandas month_name() function. Example:
>>> idx = pd.date_range(start='2018-01', freq='M', periods=3)
>>> idx
DatetimeIndex(['2018-01-31', '2018-02-28', '2018-03-31'],
dtype='datetime64[ns]', freq='M')
>>> idx.month_name()
Index(['January', 'February', 'March'], dtype='object')
For more detail visit this link.
Suppose we have a DF like this, and Date is already in DateTime Format:
df.head(3)
value
date
2016-05-19 19736
2016-05-26 18060
2016-05-27 19997
Then we can extract month number and month name easily like this :
df['month_num'] = df.index.month
df['month'] = df.index.month_name()
value year month_num month
date
2017-01-06 37353 2017 1 January
2019-01-06 94108 2019 1 January
2019-01-05 77897 2019 1 January
2019-01-04 94514 2019 1 January
using datetime object methods
I’m surpised this answer doesn’t have a solution using strftime
note, you’ll need to have a valid datetime object before using the strftime method, use pd.to_datetime(df['date_column']) to cast your target column to a datetime object.
import pandas as pd
dates = pd.date_range('01-Jan 2020','01-Jan 2021',freq='M')
df = pd.DataFrame({'dates' : dates})
df['month_name'] = df['dates'].dt.strftime('%b')
dates month_name
0 2020-01-31 Jan
1 2020-02-29 Feb
2 2020-03-31 Mar
3 2020-04-30 Apr
4 2020-05-31 May
5 2020-06-30 Jun
6 2020-07-31 Jul
7 2020-08-31 Aug
8 2020-09-30 Sep
9 2020-10-31 Oct
10 2020-11-30 Nov
11 2020-12-31 Dec
another method would be to slice the name using dt.month_name()
df['month_name_str_slice'] = df['dates'].dt.month_name().str[:3]
dates month_name month_name_str_slice
0 2020-01-31 Jan Jan
1 2020-02-29 Feb Feb
2 2020-03-31 Mar Mar
3 2020-04-30 Apr Apr
4 2020-05-31 May May
5 2020-06-30 Jun Jun
6 2020-07-31 Jul Jul
7 2020-08-31 Aug Aug
8 2020-09-30 Sep Sep
9 2020-10-31 Oct Oct
10 2020-11-30 Nov Nov
11 2020-12-31 Dec Dec
the best way would be to do with month_name() as commented by
Nurul Akter Towhid.
df[‘Month’] = df.Month.dt.month_name()
First you need to strip "0 " in the beginning (as u might get the exception leading zeros in decimal integer literals are not permitted; use an 0o prefix for octal integers)
step1)
def func(i):
if i[0] == '0':
i = i[1]
return(i)
df["Month"] = df["Month"].apply(lambda x: func(x))
Step2:
df["Month"] = df["Month"].apply(lambda x: calendar.month_name(x))